Problem: Here's a parameterization of a cylinder, for $0 < \theta < 2\pi$ : $\vec{v}(\theta, y) = (r\cos(\theta), y, r\sin(\theta))$ What is the outward-pointing vector normal to the area element of this cylinder given $r = 3$, $\theta = \dfrac{\pi}{4}$, and $y = 4$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\left( \dfrac{-3\sqrt{2}}{2}, 0, \dfrac{-3\sqrt{2}}{2} \right)$ (Choice B) B $\left( \dfrac{-3\sqrt{2}}{2}, \dfrac{-3\sqrt{2}}{2}, 0 \right)$ (Choice C) C $\left( \dfrac{3\sqrt{2}}{2}, 0, \dfrac{3\sqrt{2}}{2} \right)$ (Choice D) D $\left( \dfrac{3\sqrt{2}}{2}, \dfrac{3\sqrt{2}}{2}, 0 \right)$
Answer: The vector normal to the area element describes what's perpendicular to the surface, and the vector's magnitude represents the area of a tiny rectangle along the parameterization. After we compute it, we need to check whether it's pointing inwards or outwards. $\text{normal to area element} = \dfrac{\partial \vec{v}}{\partial \theta} \times \dfrac{\partial \vec{v}}{\partial y}$ Let's calculate the cross product. $\begin{aligned} \dfrac{\partial \vec{v}}{\partial \theta} \times \dfrac{\partial \vec{v}}{\partial y} &= \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \\ -r\sin(\theta) & 0 & r\cos(\theta) \\ \\ 0 & 1 & 0 \end{pmatrix} \\ \\ &= -r\cos(\theta) \hat{\imath} - r\sin(\theta) \hat{k} \end{aligned}$ Plugging in $r = 3$, $\theta = \dfrac{\pi}{4}$, and $y = 4$, we get the vector normal to the area element: $\begin{aligned} \dfrac{\partial \vec{v}}{\partial \theta} \times \dfrac{\partial \vec{v}}{\partial y} &= \left( -3\cos\left(\dfrac{\pi}{4}\right), 0, -3\sin\left(\dfrac{\pi}{4}\right) \right) \\ \\ &= \left( \dfrac{-3\sqrt{2}}{2}, 0, \dfrac{-3\sqrt{2}}{2} \right) \end{aligned}$ The final step is to check whether this points inwards or outwards. The cylinder defined by $\vec{v}$ has an outward facing normal vector that points directly away from the origin, but with a zero $y$ -component. Because $\theta = \dfrac{\pi}{4}$ corresponds to a positive $x$, zero $y$, and positive $z$, we should also have a positive $x$, zero $y$, and positive $z$. We need to reverse the signs of our calculation to make it match the outward normal vector. Therefore, the outward-pointing vector normal to the area element is $\left( \dfrac{3\sqrt{2}}{2}, 0, \dfrac{3\sqrt{2}}{2} \right)$.